Question

if x=2/3 and x= -3 are the roots of the equation ax^2+7x+b=0. find the value of a and b.​

Answer 1

Answer:

a(2/3)^2+7×2/3+b=0 (putting first root in the given equation)

4a/9+14/3+9b=0

4a+9b+42=0. .........(1)

Now putting 2nd root in the given equation,

a(-3)^2+7(-3)+b=0

9a+b-21=0. ........(2)

By solving eq(1) & (2) we get,

a=3

Now putting 'a' in equation (1) we get 'b' = -6

Answer 2

Given:-

• x = 2/3 and x = -3 are the roots of the equation ax² + 7x + b = 0,

To find:-

• Find the values of a and b...?

Solutions:-

• ax² + 7x + b = 0
• x = 2/3 or x = -3

We have to fine a and b.

Now,

• If x = 2/3 is a root of the equation.

=> ax² + 7x + b = 0

=> a(2/3)² + 7(2/3) + b = 0

=> 4a/9 + 14/3 + b = 0

=> (4a + 42 + 9b)/9 = 0

=> a = (-9b - 42)/4 ..........(i).

Also,

• if x = -3 is a root of the equation.

=> ax² + 7x + b = 0

=> a(-3)² + 7(-3) + b = 0

=> 9a - 21 + b = 0 .............(ii).

Now,

The multiple equation (ii). by 9 and then Subtract equation (i). we get.

=> 81a + 9b - 189 - 4a - 9b - 42 = 0

=> 77a - 231 = 0

=> a = 231/77

=> a = 3

Now, putting the value of a in Eq. (ii).

=> 9a + b - 21 = 0

=> 9(3) + b - 21 = 0

=> 27 + b - 21 = 0

=> 6 + b = 0

=> b = - 6

Hence, the value of a is 3 and b is -6.